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4r < –12
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divide both sides by 4
i meant r < -3
\[\large{\frac{4r}{4}<\frac{-12}{4}}\]
why divide?
\[\large{\frac{\cancel{4}^1 r}{\cancel{4}^1}<\frac{\cancel{-12}^{-3}}{\cancel{4}^1}}\]
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so that we can cancel 4 there in LHS
as we want to "solve for r"
soo.. i got -3?
yes
r<-3
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